Quantcast
Channel: Active questions tagged real-analysis - Mathematics Stack Exchange
Viewing all articles
Browse latest Browse all 9362

Applying L'Hopital to the log of an expression

$
0
0

The limit below is solved using multiple methods.$$\lim_{x\to\infty} \frac{x^n}{e^x}$$

However, I am trying to solve it using the comment made below the question:

Only one application of l'Hopital's rule is necessary if you takelogarithms first.

$$\lim_{x\to\infty} \frac{x^n}{e^x}=\lim_{x\to\infty}\exp \left(\ln \frac{x^n}{e^x}\right)=\lim_{x\to\infty}\exp \left(n\ln x-x\right)$$

And now I don't know how to proceed since this is the $\infty-\infty$ case. What am I missing, or was something else meant by the comment?


Viewing all articles
Browse latest Browse all 9362


<script src="https://jsc.adskeeper.com/r/s/rssing.com.1596347.js" async> </script>